ART. 77. Find the values of x in the following equations. a+mx=bx+3ab. Transpose a and b x, mx-bx=3ab a. Separate the first member into factors, one of which is x, (m-b) x=3ab-a. In this equation, x is taken m-b times, that is, the factor m -b is the coefficient of x. Divide both members by this coefficient, 3 c+ax-bx 2. a + m =2x-b. Multiply by a+m, 3c+ax-bx=2ax-ab+2mx-bm. Transpose the terms containing x into the first member, the others into the second, and reduce, ·ax-bx-2mx= -ab-bm· signs, -3 c. Change the ax + bx+2mx=ab+bm+3 c. Separate the first. member into factors, one of which is x. (a+b+2m) x=ab+bm+3c. Hence ART. 78. In pure algebra, letters are used to represent known as well as unknown quantities. It enables us to conduct operations with greater facility, and to form rules. The results of purely algebraical solutions of problems are called general formula. The design of this section is to familiarize the learner with such solutions. 1. A and B have together $270, but B has twice as much money as A. Required the money of each. In this problem it is required to divide $270 into two parts, one of which shall be double the other. Let x dollars be A's share; then 2x dollars will be the share of B. Hence, x+2x=270. This equation gives x= $90, A's share;} Ans. 2x = $180, B's share; Now, suppose that, instead of $270, A and B have any number a of dollars, of which B has twice as much as A. In this case, the problem is to divide the number a into two parts, one of which shall be twice as great as the other. Representing the shares as before, we have The general formulæ show us that one part is and the other of the number to be divided, without reference to the particular value of that number. If, in the general formulæ, we now put $270 instead of a, we have z=270 = $90, A's share; 2x= 2.270 Particular answers. 2.90 $180, B's share; 3 Let the learner substitute other numbers for a in the general formulæ, and find the particular answers. Any number divisible by 3 will give whole numbers for the answers. 2. The sum of the ages of two brothers is 76 years, and the difference of their ages is 16 years. Required the age of each. In this problem we are to separate 76 into two parts, such that the difference of those parts shall be 16. Instead of 76, let us suppose that the number to be separated into parts is indefinite, and that it is represented by a; also, that d represents the difference of the parts, that is, the excess of the greater part above the less. Let x the less part; then z+d the greater part. Hence, x+x+d=a. Reduce and transpose d, To obtain the greater part, we add d to the less, and we have α x + d = 2/2/2 =-2+d. Change d in the 2d member to α halves, a x+d==+==a1a, the greater part. 2 2 2 By examining the general formulæ for the two parts, and recollecting that a and d mày stand for any numbers, provided d is less than a, we see that they furnish the following rule for separating a number into two parts, whose difference is given. The less part is found by subtracting half of the difference of the parts from half of their sum, that is, from half of the number to be separated; or, by subtracting the difference of the parts from their sum, and dividing the remainder by 2. The greater part is found by adding half of the difference to half of the sum of the parts; or, by adding the difference to the sum of the parts, and dividing the amount by 2. Let the learner separate the following numbers into parts, either by means of the formulæ, or by following he rule. 10. A man bought corn at a shillings, rye at b shillings, and wheat at c shillings, per bushel, and the whole amounted to a shillings. Required the number of bushels of each he bought, if he bought the same number of bushels of each. Let x the number of bushels of each. Then, Separating the 1st member into factors, one of which is x, (Art. 59.) (a+b+c) x=d; dividing by a+b+c, the coefficient x= d of x, the number of bushels of each. This general formula may be translated into the following rule. Divide the price of the whole by the sum of the prices of a bushel of each sort; the quotient will be the number of bushels of each. If, in the formula, we substitute 5, 6, 7, and 180, for a, b, c, and d, respectively, we have x= 180 180 5+6+7 18 10 bushels, particular answer. This rule may be extended to any number of articles; only we must change "bushel" and "number of bushels into other corresponding expressions, as the case may require. 11. How many apples at 1 cent, pears at 2 cents, peaches at 3 cents, and oranges at 4 cents, apiece, of each an equal number, can I buy for $5? 12. In a certain manufactory there are employed b |